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ahrayia [7]
3 years ago
6

A civil engineer is analyzing the compressive strength of concrete. Compressive strength is normally distributed with A random s

ample of 12 sample specimens has a mean compressive strength of psi. Round your answers to 1 decimal place. (a) Calculate the 95% two-sided confidence interval on the true mean compressive strength of concrete. Enter your answer; 95% confidence interval, lower bound Enter your answer; 95% confidence interval, upper bound (b) Calculate the 99% two-sided confidence interval on the true mean compressive strength of concrete.
Mathematics
1 answer:
wel3 years ago
3 0

Answer:

95%: (3278.354 ; 3270.083)

99% : (3221.646 ; 3278.354)

Step-by-step explanation:

Given :

Sample size, n = 12

Mean, xbar = 3250

Sample standard deviation = √1000

The 95% confidence interval :

Mean ± Margin of error

Margin of Error = Tcritical * s/√n

Tcritical at 0.05, df=12-1 = 11 ;

Tcritical at 95% = 2.20

Hence,

Margin of Error = (2.20 * √1000/√12) = 20.083

Confidence interval : 3250 ± 20.083

Lower boundary = 3250 - 20.083 = 3229.917

Upper boundary = 3250 + 20.083 = 3270.083

2.)

The 99% confidence interval :

Mean ± Margin of error

Margin of Error = Tcritical * s/√n

Tcritical at 0.01, df=12-1 = 11 ;

Tcritical at 99% = 3.106

Hence,

Margin of Error = (3.106 * √1000/√12) = 28.354

Confidence interval : 3250 ± 28.354

Lower boundary = 3250 - 28.354 = 3221.646

Upper boundary = 3250 + 28.354 = 3278.354

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Answer:

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Step-by-step explanation:

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Adding school fairest 20% of the people are under 16 years old one third of the people remaining are teachers if there are 21 te
dedylja [7]
<span>In the fairest school 20% are below 16 years old
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4 0
3 years ago
6.
larisa86 [58]

A. Every month Population will increase by a factor of 0.84%.

B.  Every 3 months Population will increase by a factor of 2.5%.

C. Increase in population in every 20 months is 10% + 6.72% = 16.72%.

<u>Step-by-step explanation:</u>

Here, we have  number of employees in a company has been growing exponentially by 10% each year. So , If we have population as x in year 2019 , an increase of 10% in population in 2020 as  x + \frac{x}{100}\times10 which is equivalent to \frac{11x}{10}.

<u>A.</u>

For each month: We have 12 months in a year and so, distributing 10% in 12 months would be like \frac{10}{12}  = 0.84 . ∴ Every month Population will increase by a factor of 0.84%.

<u>B.</u>

In every 3 months: We have , 12 months in a year , in order to check for every 3 months  \frac{12}{3} = 4 and Now, Population increase in every 3 months is \frac{10}{4}  = 2.5.  ∴ Every 3 months Population will increase by a factor of 2.5%.

<u>C.</u>

In every 20 months: We have , 12 months in a year in which increase in population is 10% . Left number of moths for which we have to calculate factor of increase in population is 20-12 = 8. For 1 month , there is 0.84% increase in population ∴ For 8 months , 8 × 0.84 = 6.72 %.

So , increase in population in every 20 months is 10% + 6.72% = 16.72%.

5 0
3 years ago
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